A right circular cone sits on a table, pointing up. The cross-section triangle, perpendicular to the base, has a vertex angle of 60 degrees. The diameter of the cone's base is $12\sqrt{3}$ inches. A sphere is placed inside the cone so that it is tangent to the sides of the cone and sits on the table. What is the volume, in cubic inches, of the sphere? Express your answer in terms of $\pi$.
Solution: Since the vertex angle of the cross-section triangle measures $60^\circ$, the cross-section triangle is equilateral.  Also, the cross-section of the sphere inscribed in the cone is a circle tangent to each of the triangle's sides.  Call the vertices of the equilateral triangle $A$, $B$, and $C$, and let $O$ be the center of the circle and $D$ and $E$ the midpoints of segments $AB$ and $BC$, respectively.  To find the radius of the circle, divide the 30-60-90 triangle $CDB$ into three smaller congruent 30-60-90 triangles as shown.  Since the area of each of these triangles is smaller by a factor of $3$ than the area of triangle $CDB$, each corresponding side must be smaller by a factor of $\sqrt{3}$.  Thus $OE=DB/\sqrt{3}=6$ inches.  Therefore, the volume of the sphere is $V=\frac{4}{3}\pi(\text{radius})^3=\frac{4}{3}\pi(\text{6 inches})^3=\boxed{288\pi}$ cubic inches.

[asy]

size(2.5inch);

pair A = (0,0);
pair B = (2,0);
pair C = (1,sqrt(3));
pair O = (1,sqrt(3)/3);

draw(O--B);
draw(O--C);
draw(O--(B+C)/2);
draw(O--(A+B)/2);

draw(A--B--C--cycle);
draw(circle(O,sqrt(3)/3));

label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",(A+B)/2,S);
label("$E$",(B+C)/2,NE);
label("$O$",O,W);
label("$12\sqrt{3}$",(1,-0.2),S);

[/asy]